JEE Mains 2026: Most Scoring Topics in Chemistry, Chapter-wise Weightage

Duma's Method - Practice Questions & MCQ

Edited By admin | Updated on Sep 18, 2023 18:35 AM | #JEE Main

Syllabus

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Duma's Method is considered one the most difficult concept.
7 Questions around this concept.

Solve by difficulty

EASY

Given below are two statements I and II.
Statement I : Dumas method is used for estimation of “Nitrogen” in an organic compound.
Statement II : Dumas method involves the formation of ammonium sulphate by heating the organic compound with conc H2SO4.
In the light of the above statements, choose the correct answer from the options given below

Both Statement I and Statement II are true

Statement I is false but Statement II is true

Both Statement I and Statement II are false

Statement I is true but Statement II is false

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Concepts Covered - 1

Duma's Method

The nitrogen containing organic compound, when heated with copper oxide in an atmosphere of carbon dioxide, yields free nitrogen in addition to carbon dioxide and water.

${\mathrm{C}_{\mathrm{x}} \mathrm{H}_{\mathrm{y}} \mathrm{N}_{\mathrm{z}}+(2 \mathrm{x}+\mathrm{y} / 2) \mathrm{CuO} \longrightarrow} {\: \mathrm{x} \mathrm{CO}_{2}+\mathrm{y} / 2 \mathrm{H}_{2} \mathrm{O}+\mathrm{z} / 2 \mathrm{N}_{2}+(2 \mathrm{x}+\mathrm{y} / 2) \mathrm{Cu}}$
Traces of nitrogen oxides formed, if any, are reduced to nitrogen by passing the gaseous mixture over a heated copper gauze. The mixture of gases so produced is collected over an aqueous solution of potassium hydroxide which absorbs carbon dioxide. Nitrogen is collected in the upper part of the graduated tube as shown in the figure.
Let the mass of organic compound = mg
Volume of nitrogen collected = V₁ mL
Room temperature = T₁K

$\text { Volume of nitrogen at } \mathrm{STP}=\frac{p_{1} V_{1} \times 273}{760 \times T_{1}}$
Where p₁ and V₁ are the pressure and volume of nitrogen, p₁ is different from the atmospheric pressure at which nitrogen gas is collected. The value of p₁ is obtained by the relation;
p₁ = Atmospheric pressure – Aqueous tension
22400 mL N₂ at STP weighs 28g.

$\begin{array}{l}{V \mathrm{\: mL\: } \mathrm{N}_{2} \text { at STP weighs }=\frac{28 \times V}{22400} \mathrm{g}} \\\\ {\text {Percentage of nitrogen }=\frac{28 \times V \times 100}{22400 \times m}}\end{array}$